See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: E2 reaction rate depends on (i) the leaving group ability, (ii) the base strength/nucleophilicity, and (iii) the substrate structure (degree of substitution and orbital overlap/anti-periplanar geometry). Pair 1 vs 2 (leaving group): Reaction (1) uses an alkyl chloride (Cl^- leaving group) while reaction (2) uses an alkyl iodide (I^- leaving group). I^- is a much better leaving group than Cl^- because C-I bond is weaker and iodide is more stable as a leaving group. Therefore, reaction (2) has a greater rate constant than reaction (1). Pair 3 vs 4 (base/nucleophile): Both involve 2-bromopropane as the substrate. Reaction (3) uses CH3O^- (methoxide) and reaction (4) uses CH3S^- (methanethiolate). For E2, a stronger base gives a faster rate. Methoxide (CH3O^-) is a stronger base than methanethiolate (CH3S^-) because oxygen is more electronegative and holds negative charge less stably than sulfur (sulfur is larger, more polarizable, and better at stabilizing negative charge, making it a weaker base). Therefore, methoxide is a stronger base, and reaction (3) has the greater rate constant. So from pair 3/4, reaction (3) is faster. Pair 5 vs 6 (substrate structure): Reaction (5) involves a substrate where bromine is on a non-benzylic secondary carbon (the beta carbon to the phenyl group), while reaction (6) involves bromine directly on the benzylic carbon. For E2 elimination, the transition state is stabilized when the developing double bond is conjugated with the phenyl ring. In reaction (5), elimination would form a double bond between the benzylic carbon and the adjacent carbon, giving a styrene-type product (conjugated with phenyl), whereas in reaction (6), with Br at the benzylic position, elimination would also give a styrene-type product. Re-examining: In reaction (5), the substrate is PhCH2-CH(Br)-CH3; elimination of HBr gives PhCH=CHCH3 (conjugated, stabilized). In reaction (6), the substrate is PhCH(Br)-CH2CH3; elimination gives PhCH=CHCH3 or Ph-C(=CH2)-... Actually in (5) the Br is NOT at benzylic position so the anti-periplanar arrangement and transition state stabilization differ. The key point: reaction (5) has Br on a secondary carbon beta to phenyl — the transition state for E2 is stabilized by partial conjugation with the phenyl ring as the double bond forms. Reaction (6) has Br at the benzylic position; while benzylic C-Br bonds are weaker, the E2 transition state leading away from phenyl is less stabilized. Overall, reaction (5) proceeds faster due to better transition state stabilization through conjugation. Therefore reaction (5) has the greater rate constant. Summary: The reactions with greater rate constants are (2), (3), and (5), corresponding to answer choice (c). Why other options fail: - (a) 2, 4, 6: Incorrect because (3) is faster than (4) and (5) is faster than (6). - (b) 1, 3, 5: Incorrect because (2) is faster than (1) due to better leaving group. - (d) 2, 4, 5: Incorrect because (3) is faster than (4) since methoxide is a stronger base than methanethiolate. Therefore, the correct answer is C.